operator = problem. Why is this happening? [closed] - c++

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So i have written the following method in a class:
Array & operator = (const Array &a) {
/*delete [] array;
priv_size = a.priv_size; HOW DOES THIS EVEN DO SOMETHING???
priv_base = a.priv_base;
array = new int[priv_size*priv_size];
for (int i=0; i<priv_size*priv_size; i++)
array[i] = a.array[i];
return *this;
*/
}
as you can clearly see it is all in one big comment. However, when i call the = operator in main like this:
cout << a; //there is an operator << too
cout << endl;
Array b = a;
b[1][1] = 39;
b[3][1] = 2;
cout << b;
cout << endl;
cout << a;
where a is the following array:
0 0 0 0
0 0 0 0
3 0 0 0
0 0 0 5
this happens:
0 0 0 0
0 0 0 0
3 0 0 0
0 0 0 5
0 0 0 0
0 39 0 0
3 0 0 0
0 2 0 5
0 0 0 0
0 0 0 0
3 0 0 0
0 0 0 5
As you can see i also change some values in b, but a remains intact which means b is not by mistake somehow connected to a in any way.
What is going on here? Help.
Thanks.
EDIT: I moved the comment so that it covers the whole method including the Array & operator = (const Array &a) part. The exact same thing happens(????). How is this possible? It almost seems like there is already a default operator = method, which does't really make too much sense, because why would c++ prioritize such a method, and even if it does, why set it so that it copies a whole array?.
Also, here is the constructor, since a lot of people needed to see it:
Array (int s = 0, int b = 0) :
array(new int[s*s]), priv_size(s), priv_base(b) {
for (int i=0; i<priv_size*priv_size; i++)
array[i] = 0; //sets everything to 0
}

Answer 1

My question is answered:
In my example i wrote Array b = a;, which in c++ calls the copy constructor and not the opperator =
This is documented here:
https://en.cppreference.com/w/cpp/language/copy_initialization
As well as here:
https://en.cppreference.com/w/cpp/language/copy_constructor
Thanks to everyone who took the time to read the question and comment.